How to use mean theorem to solve this proof

[transgalactic]

f(x) continues in [a,b] interval,and differentiable (a,b) b>a>0
alpha differs 0
proove that there is b>c>a

in that formula:
http://img392.imageshack.us/my.php?image=81208753je3.gif

my trial:
i mark alpha as "&"
mean theorim says f'(c)=[(f(b)-f(a)]/(b-a)

[f(b)*(a^&) - f(a)*b^&] / [a^& - b^&]= f(c) -c * [(f(b)-f(a)]/[(b-a) * &]using mean theorem i replaced f'(c)
what should i do next??

 

[HallsofIvy]

You want to use the mean value theorem to prove that, for appropriate conditions on f, a, b, and $\alpha$ that you don't give,
$$\frac{\left|\begin{array}{cc}f(a) & a^\alpha \\ f(b) & b^\alpha\end{array}\right|}{a^\alpha- b^\alpha}= f(c)- \frac{cf'(c)}{\alpha}$$

That is the same as
[tex]\frac{a^\alpha f(b)- b^\alpha f(a)}{a^\alpha- b^\alpha}= f(c)- \frac{cf(c)}{\alpha}[/itex]

Obviously we have to construct some function $\phi$ to which to apply the mean value theorem. My first thought was something like $x^\alpha f(x)$ but I notice that "a" and "b" are swapped in the numerator. The line y= b-(x-a)= a+b- x passes thorugh (a,b) and (b,a) so something like $x^\alpha f(b+a-x)$ should work.

But then we would be applying the mean vaue theorem at a and b and the denominator a- b not $a^\alpha- b^\alpha$. That means we must be applying the mean value theorem at $a^\alpha$ and $b^\alpha$ so need to "fix" the argument of f to give b when $b= a^\alpha$ and vice- versa.

That finally gives us $\phi(x)= xf((a^\alpha+ b^\alpha- x) ^{1/\alpha})$. The mean value theorem, applied to $\phi$ on the interval from $a^\alpha$, $b^\alpha$ gives
[tex]\frac{\phi(a^\alpha)- \phi(b^\alpha)}{a^\alpha- b^\alpha}= \phi'(c')[/itex]
for some c' between $a^\alpha$ and $b^\alpha$. c' here is not the "c" in you original formula.

 

[transgalactic]

i thought we apply mean theorim to f(x)

its very similar and we have f'(c) which is the same as the original fornula
??

 

[Pere Callahan]

i thought we apply mean theorim to f(x)

its very similar and we have f'(c) which is the same as the original fornula
??

If you had known how to apply the MVT to that question you would have done so, wouldn't you? SInce you opened this thread and asked how to prove that fancy equation using the MVT I assume you didn't know how to beneficially apply the theorem. (To whatever function, be it f itself or be it that some auxiliary function needs to be constructed.)

So, I suggest you try and understand what HallsofIvy wrote; only expressing your awe in view of some more elaborate reasoning than your own is little constructive.

 

[transgalactic]

You want to use the mean value theorem to prove that, for appropriate conditions on f, a, b, and $\alpha$ that you don't give,
$$\frac{\left|\begin{array}{cc}f(a) & a^\alpha \\ f(b) & b^\alpha\end{array}\right|}{a^\alpha- b^\alpha}= f(c)- \frac{cf'(c)}{\alpha}$$

That is the same as
[tex]\frac{a^\alpha f(b)- b^\alpha f(a)}{a^\alpha- b^\alpha}= f(c)- \frac{cf(c)}{\alpha}[/itex]

Obviously we have to construct some function $\phi$ to which to apply the mean value theorem. My first thought was something like $x^\alpha f(x)$ but I notice that "a" and "b" are swapped in the numerator. The line y= b-(x-a)= a+b- x passes thorugh (a,b) and (b,a) so something like $x^\alpha f(b+a-x)$ should work.

But then we would be applying the mean vaue theorem at a and b and the denominator a- b not $a^\alpha- b^\alpha$. That means we must be applying the mean value theorem at $a^\alpha$ and $b^\alpha$ so need to "fix" the argument of f to give b when $b= a^\alpha$ and vice- versa.

That finally gives us $\phi(x)= xf((a^\alpha+ b^\alpha- x) ^{1/\alpha})$. The mean value theorem, applied to $\phi$ on the interval from $a^\alpha$, $b^\alpha$ gives
[tex]\frac{\phi(a^\alpha)- \phi(b^\alpha)}{a^\alpha- b^\alpha}= \phi'(c')[/itex]
for some c' between $a^\alpha$ and $b^\alpha$. c' here is not the "c" in you original formula.

"y= b-(x-a)= a+b- x "
whats the role of this??